Summary
The contract that Lazer company for a part of Boeing corporation has to have a diameter average equal to 6 inches and a standard deviation of 0.10 inches. Below are discussions and explanations to help the Lazer company with what they should conclude, given the mean of the diameter sample for 200 parts as 6.3 inches. The constructed null and alternative hypotheses should have the correct parameters and the developed decision rule that assumes the sample size to be 200 parts and their significance level to be 0.01.
At this level of significance of 0.0, much evidence concludes that the mean diameter of the parts given by the test run cannot be 6 inches. Hence Lazer company should conclude that the mean diameter of the parts from the testing run is different from the requirement in the contract, which is 6 inches (Light, 2018). To find the conclusion, we will have to see if the null hypothesis is true and if not, then consider the alternative hypothesis to check if it gives the required answer. Note the following:
- σ, which is the standard deviation
- x is the individual data value
- µ, which is the mean
- In this:
- Make X as the diameter of parts given by the test that is run and the µ0 = 6, σ= 0.10
The null hypothesis, which states that the “mean diameter of parts produced by test run is 6 inches,” will be:
- H0: µ is to be 6 inches
The alternative hypothesis states that the “mean diameter of parts produced by test run is not equal to 6 inches”; hence one has to calculate to prove that the mean diameter can’t be 6 inches. These will be:
- Ha: µ≠6 inches
Given
Following the central limit theorem, the sample will follow a normal distribution, consisting of the curves uniquely defined by the mean and standard deviation in a given population because the sampling size is large. Therefore, the decision rule will then be:
- Reject H0 if |z| >za /2 since α=0.01, we have to get the z0.005. To make this possible, we have to use a function called NORM.INV has (probability, mean, and standard) in that Probability is 1-0.05=0.995, using the z-distribution, which is a normal distribution it tells one the standard deviations away an individual data value has fallen from the mean, the mean, which is 0 and the standard which is 1
- Hence the NORM.INV will be (0.995, 0, 1) which produces 2.575829304.
- Hence decision rule will be: Reject H0 if |z| > 2.575829304
x =6.03 inches and n=200, where n is the sample size of the parts
We need to solve for the test statistic z using the formula:
- z= x -µ0. Hence we will have z= 6.03-6
- σ/ √n 0.10/√200
To solve for this, we will have to encode = (6.03-6)/ (0.10/SQRT (200)) in excel.
Hence, we will have z=4.242640687.
From the decision rule, since |z=4.242640687| > 2.575829304, it rejects the null hypothesis, i.e., the null hypothesis is incorrect, making the alternative hypothesis correct.
This shows that at 0.01 which is a significant level sufficient to support the conclusion that the mean diameter of the parts stated by the testing is not 6 inches. Therefore, Lazer company should conclude that the mean diameter of the given parts from the test run is different from the requirement in the contract, which is 6 inches.
Reference
Light, G. L. (2018). On the mean of sample-standard-deviation. International Journal of Applied Mathematics, 31(3), 359-370. Web.